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Hence, the comparison of solid angles becomes a matter of great ease and simplicity : for, since the areas of spherical triangles are measured by the excess of the sums of their angles each above two right angles (th. 5); and the areas of spherical polygons of n sides, by the excess of the sum of their angles above 2n – 4 right angles (th. 6); it follows, that the magnitude of a trilateral solid angle, will be measured by the excess of the sum of the three angles, made respectively by its bounding planes, above 2 right angles; and the mag. nitudes of solid angles formed by n bounding planes, by the excess of the sum of the angles of inclination of the several planes above 2n -- 4 right angles.

As to solid angles limited by curve surfaces, such as the angles at the vertices of cones; they will manifestly be measured by the spheric surfaces cut off by the prolongation of their bounding surfaces, in the same manner as angles determined by planes are measured by the triangles or polygons, they mark out upon the same, or an equal sphere. In all cases, the maximum limit of solid angles, will be the plane towards which the various planes determining such angles approach, as they diverge further from each other about the same summit: just as a right line is the maximum limit of plane angles, being formed by the two bounding lines when they make an angle of 180°. The maximum limit of solid angles is measured by the surface of a hemisphere, in like manner as the maximum limit of plane angles is measured by the arc of a semicircle. The solid right angle (either angle, for example, of a cube) is (= ?) of the maximum solid an, gle: while the plane right angle is half the maximum plane angle.

The analogy between plane and solid angles being thus traced, we may proceed to exemplify this theory by a few in'stances; assuming 1000 as the numeral measure of the maxi, mum solid angle = 4 times 90° solid = 360° solid.

1. The solid angles of right prisms are compared with great facility. For, of the three angles made by the three planes which, by their meeting, constitute every such solid angle, two are right angles; and the third is the same as the corre, sponding plane angle of the polygonal base; on which, therefore, the measure of the solid angle depends. Thus, with respect to the right prism with an equilateral triangular base, each solid angle is formed by planes which repectively make angles of 90°, 90°, and 60°. Consequently 90°+90° +60°180°=60°, is the measure of such angle, compared with 360° the maximum angle. It is, therefore, one-sixth of the maxinum angle. A right prism with a square base, has, in like


square base


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2 m

manner, each solid angle measured by 90° +90° + 90°- 180° =90°, which is of the maximum angle. And thus it may be found, that each solid angle of a right prism, with an equilateral triangular base is & max. angle =

= 3.1000. is

g .1000. pentagonal base is

o.1000. hexagonal is į

th:1000. heptagonal is

= .1000. octagonal is

.1000, nonagonal is

1.1000. decagonal is

&.1000. undecagonal is

=a.1000. duodecagonal is it

000. m gonal is

.1000. Hence it may be deduced, that each solid angle of a reguIar prism, with triangular base, is half each solid angle of a prism with a regular hexagonal base. Each with regular square base = of each, with regular octagonal base,

= pentagonal = 2

decagonal, hexagonal = 5

duodecagonal, in gonal

m gonal base. Hence again we may infer, that the sum of all the solid angles of any prism of triangular base, whether that base be regular or irregular, is half the sum of the solid angles of a prism of quadrangular base, regular or irregular. And, the sum of the solid angles of any prism of tetragonal base is = sum of angles in prism of pentag. base, pentagonal

hexagonal, hexagonal

heptagonal, ni gonal

(m+1) gonal, 2. Let us compare the solid angles of the five regular bodies. In these bodies, if m be the number of sides of each face; n the number of planes which meet at each solid angle; 10 = half the circumference or 180°; and A the plane angle

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m - 2




made by two adjacent faces: then we have sin ja=

1 sin


This theorem gives, for the plane angle formed by every two contiguous faces of the tetraëdron, 70° 31'42"; of the hexaédron, 90°; of the octaëdron, 109°28'18"; of the dodecaëdron, 116°33'54"; of the icosaëdron, 138°11'23". But, in these polyedræ, the number of faces meeting about each solid angle,

3, 3, subdivision

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3, 3, 4, 3, 5 respectively. Consequently the several solid angles will be determined by the subjoined proportions :

Solid Angle. 360°: 3.70°31'42" -- 180° :: 1000 : 87.73611 Tetraëdron. 360° : 3.90° - 180° :: 1000 : 250• Hexaëdron. 360° : 4.109° 28'18" – 360°:: 1000 : 216.35185 Octaëdron. 360° : 3.116°33'54" - 180° :: 1000 : 471•395 Dodecaëdron. 360° : 5.138°11'23" – 540° :: 1000 : 419.30169 Icosaëdron,

3. The solid angles at the vertices of cones, will be determined by means of tlie spheric segments cut off at the bases of those cones; that is, if right cones, instead of having plane bases, had bases formed of the segments of equal spheres, whose centres were the vertices of the cones, the surfaces of those segments would be measures of the solid angles at the respective vertices. Now, the surfaces of spheric segments, are to the surface of the hemisphere, as their altitudes, to the radius of the sphere; and therefore the solid angles at the vertices of right coness will be to the maximum solid angle, as the excess of the slånt side above the axis of the cone, to the slant side of the cone. Thus, if we wish to ascertain the solid angles at the vertices of the equilateral and the rightangled cones; the axis of the former is įV3, of the latter, İV2, the slant side of each being unity. Hence,

Angle at vertex. 1:1-1V3 :: 1000 : 133.97464, equilateral cone, 1:1-ŽV2 :: 1000 : 292.89322, right-angled cone.

4. From what has been said, the mode of deterinining the solid angles at the vertices of pyramids will be sufficiently obvious. If the pyramids be regular ones, if n be the number of faces meeting about the vertical angle in one, and a the angle of inclination of each two of its plane faces; if x be the number of planes meeting about the vertex of the other, and a the angle of inclination of each two of its faces : then will the vertical angle of the former, be to the vertical angle of the latter pyramici, as NA-(N-2) 180°, to na--(n-2) 180°

If a cube be cut by diagonal planes, into 6 equal pyramids with square bases, their vertices all meeting at the centre of the circumscribing sphere; then each of the solid angles, made by the four planes meeting at each vertex, will be of the maximum solid angle; and each of the solid angles at the bases of the pyramids, will be of the maximum solid angle. Therefore, each solid angle at the base of such

pyramid, is one-fourth of the solid angle at its vertex: and, if the angle at the vertex be bisected, as described below, either of the solid angles arising from the bisection, will be double of either solid angle at the base. Hence also, and from the first

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any two

subdivision of this scholium, each solid angle of a prism, with equilateral triangular base, will be half each vertical angle of these pyramids, and double each solid angle at their bases.

The angles made by one plane with another, must be ascertained, either by measurement or by computation, according to circumstances. But, the general theory being thus explained, and illustrated, the further application of it is left to the skill and ingenuity of geometers; the following simple example, merely, being added here.

Er. Let the solid angle at the vertex of a square pyramid be bisected.

1st. Let a plane be drawn through the vertex and opposite angles of the base, that plane will bisect the solid angle at the vertex; forming two trilateral angles, each equal to half the original quadrilateral angle.

2dly. Bisect either diagonal of the base, and draw any plane to pass through the point of bisection and the vertex of the pyramid; such plane, if it do not coincide with the former, will divide the quadrilateral solid angle into two equal quadrilateral solid angles. For this plane, produced, will bisect the great circle diagonal of the spherical parallelogram cut off by the base of the pyramid ; and any great circle bisecting such diagonal is known to bisect the spherical parallelogram, or square; the plane, therefore, bisects the solid angle.

Cor. Hence an indefinite number of planes may be drawn, each to bisect a given quadrilateral solid angle.


Resolution of Spherical Triangles.

The different cases of spherical trigonometry, like those in plane trigonometry, may be solved either geometrically or algebraically. We shall here adopt the analytical method, as well on account of its being more compatible with brevity, as because of its correspondence and connexion with the substance of the preceding chapter*. The whole doctrine may Þe comprehended in the subsequent problems and theorems.

For the geometrical method, the reader may consult Simson's or Playfair's Euclid, or Bishop llursley's Elementary Treatises on Practical Mathematics.


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To Find Equations, from which may be deduced the Solution of all the Cases of Spherical Triangles.

Let ABC be a spherical triangle; AD the tangent, and GD the secant, of the are AB; AE the tangent, and GE the se

cant, of the arc AC; let

the capital letters A, B, C,

denote the angles of the

triangle, and the small letters a, b, c, the op

posite sides BC, AC, AB.

Then the first equations in art. 6 PI. Trig.

applied to the two triangles ADE, GDE, give, for the former, DE tan b + tan2 e — tan b . tán c. cos A; for the latter, DE2 :sec2 b + sec2 c sec b. sec c. cos a. Substracting the first of these equations from the second, and observing that sec2 b tan2b=R2


sin b. cos c
cos b. cos c

1, we shall have, after a little
0. Whence


cos a
cos b. cos c

reduction, 1+
the three following symmetrical equations are obtained:
cos a = cos b. cos c + sin b. sin e. cos A
cos b
= cos a. cos e + sin a. sin c COS B
cos c = còs a . cos b + sin a. sin b, cos c





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In Every Spherical Triangle, the Sines of the Angles are Proportional to the Sines of their Opposite Sides.


If, from the first of the equations marked 1, the value of cos a be drawn, and substituted for it in the equation sin A 1. cos A, we shall have

cos2 a + cos2 b. cos2 c - 2 cos a cos è, cos c
sin2 b. sin2 c

sin' A = 1
Reducing the terms of the second side of this equation to a
common denominator, multiplying both numerator and deno-
minator by sin2 a, and extracting the sq. root there will result
√(1-cos2 a- cos2 b— cos2 c + 2 cos a. cos b. cos c)

sin a.

sin b. sin c

sin A sin a.
Here, if the whole fraction which multiplies sin a, be denoted
by K (see art. 8 chap. iii), we may write sin A = K. sin a.
And, since the fractional factor, in the above equation, con-
tains terms in which the sides a, b, c, are alike affected, we

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